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C = 2+2^2 + 2^3 + 2^4 + 2^5 ..... + 2^97 + 2^98 + 2^99 + 2^100
= ( 2 + 2^2 + 2^3 + 2^4 + 2^5)+........+ (2^97 + 2^98 + 2^99 + 2^100 )
= 2(1+2+2^2 + 2^3 + 2^4)+..........+ 2^96(1+2+2^2 + 2^3 + 2^4)
=2.31 +...........+2^96.31
= 31(2+....+2^96)
Vì 31(2+....+2^96) chia hết cho 31
nên C chia hết cho 31
Nhớ bấm đúng cho mình nhé bạn!!!!!!!!
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A=5+52+...+599+5100
=(5+52)+...+(599+5100)
=5.(1+5)+...+599.(1+5)
=5.6+...+599.6
=6.(5+...+599) chia hết cho 6 (dpcm)
Ccá câu khcs bạn cứ dựa vào câu a mà làm vì cách làm tương tự chỉ hơi khác 1 chút thôi
Chúc bạn học giỏi nha!!
\(A=5+5^2+5^3+...+5^{100}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...\left(5^{99}+5^{100}\right)\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{99}\left(1+5\right)\)
\(=5.6+5^3.6+...+5^{99}.6\)
\(=6\left(5+5^3+...+5^{99}\right)⋮6\)(đpcm)
\(B=2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+...+2^{96}.31\)
\(=31\left(2+...+9^{96}\right)⋮31\)(đpcm)
\(C=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)
\(=3.4+3^3.4+...+3^{59}.4\)
\(=4\left(3+3^3+...+3^{59}\right)⋮4\)(đpcm)
\(C=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)
\(=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
\(=3.13+...+3^{58}.13\)
\(=13\left(3+...+3^{58}\right)⋮13\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
C = 2 + 22 + 23 + ...+ 2100
Ta chia C thành 20 nhóm 5 số hạng liên tiếp:
C = (2 + 22 + 23 + 24 + 25) + (26 + 27 + 28 + 29 + 210) + ...+ (296 + 297+ 298 + 299 + 2100)
= 2 (1 + 2 + 22 + 23 + 24) + 26(1 + 2 + 22 + 23 + 24) + ...+ 296(1 + 2 + 22 + 23 + 24)
= (1 + 2 + 22 + 23 + 24) (2 + 26 + ...+ 296)
= (1 + 2 + 4 + 8 + 16). (2 + 26 + ...+ 296)
= 31. (2 + 26 + ...+ 296)
Dễ thấy C chia hết cho 31.
Chứng minh rằng : C = 2 + 2^2 + 2 + 3 + .......... + 2^99 + 2^100 chia hết cho 31 . Và tính tổng C .
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\(A=\left(2+2^2\right)+...+\left(2^{99}+2^{100}\right)\)
\(A=2\cdot\left(1+2\right)+...+2^{99}\cdot\left(1+2\right)\)
\(A=2\cdot3+...+2^{99}\cdot3\)
\(A=3\cdot\left(2+...+2^{99}\right)⋮3\left(đpcm\right)\)
2 ý kia tương tự
Giải:
Đặt S=(2+2^2+2^3+...+2^100)
=2.(1+2+2^2+2^3+2^4)+2^6.(1+2+2^2+2^3+2^4)+...+(1+2+2^2+2^3+2^4).296
=2.31+26.31+...+296.31
=31.(2+26+...+296)\(⋮\)31
![](https://rs.olm.vn/images/avt/0.png?1311)
lg
a)C=3+3^2+3^3+...+3^100
=(3+3^2+3^3+3^4)+...+(3^96+3^97+3^98+3^99+3^100)
=(3.1+3.3+3.3^2+3.3^3)+...+(3^96.1+3^96.3+3^96.3^2+3^96.3^3)
=3.(1+3+3^2+3^3)+...+3^96.(1+3+3^2+3^3)
=3.40+...+3^96.40
=40.(3+...+3^96) chia hết cho 40
=>C chia hết cho 40
Vậy C chia hết cho 40
phần b làm tương tự
a, sai đề
b,Ta có :
C=2+2^2+2^3+2^4+2^5...+2^96+2^97+2^98+2^99+2^100
= (2+2^2+2^3+2^4+2^5)+...+(2^96+2^97+2^98+2^99+2^100)
= (2.1+2.2+2.2^2+2.2^3+2.2^4)+...+(2^96.1+2^96.2+2^96.2^2+2^96.2^3+2^96.2^4)
=2. (1+2+2^2+2^3+2^4) +...+2^96.(1+2+2^2+2^3+2^4)
=2.31+...+2^96.31
=31. (2+...+2^96) chia hết cho 31
=>C chia hết cho 31
![](https://rs.olm.vn/images/avt/0.png?1311)
a. C = 2 + 22 + 23 + …….. + 299 + 2100
= 2(1 +2 + 22+ 23+ 24) + 26(1 + 2 + 22+ 23+ 24)+…+ (1 + 2 + 22+ 23+ 24).296
= 2 . 31 + 26 . 31 + … + 296 . 31 = 31(2 + 26 +…+296).
Vậy C chia hết cho 31
b. C = 2 + 22 + 23 + …….. + 299 + 2100 à 2C = 22 + 23 + 24 + …+ 2100 + 2101
Ta có 2C – C = 2101 – 2 \(\Rightarrow\) 2101 = 22x-1 \(\Rightarrow\)2x - 1 = 101
2x = 102
=> x = 51
\(C=2+2^2+...+2^{100}\)
\(C=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^5+2^6+2^7+2^9+2^{10}\right)...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(C=2.\left(1+2+2^2+2^3+2^4\right)+2^5\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(C=2.31+2^5.31+...+2^{96}.31\)
\(C=31.\left(2+2^5+...+2^{96}\right)\)(VÌ TÍCH CÓ THỪA SỐ 31 NÊN CHIA HẾT CHO 31)